Find the zeros of the function. Enter the solutions from least to greatest. $f(x) = (x + 3)^2 - 4$ $\text{lesser }x = $
Explanation: $\begin{aligned} (x + 3)^2 - 4&= 0 \\\\ (x+3)^2&=4 \\\\ \sqrt{(x+3)^2}&=\sqrt{4} \end{aligned}$ $\begin{aligned} x+3&=\pm2 \\\\ x&=\pm2-3 \\ \phantom{(x + 3)^2 - 4}& \\ x=-5&\text{ or }x=-1 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= -5 \\\\ \text{greater } x &= -1 \end{aligned}$